a) ĐKXĐ: \(x>0,x\ne1\)
\(B=1:\dfrac{\left(x+2\right)\left(\sqrt{x}+1\right)+\left(\sqrt{x}+1\right)\left(x-1\right)-\left(\sqrt{x}+1\right)\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x\sqrt{x}-\sqrt{x}}\)
\(=\dfrac{\left(x-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(x-1\right)}=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}\)
b) \(B=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}=\sqrt{x}+1+\dfrac{1}{\sqrt{x}}\)
Áp dụng BĐT Cauchy cho 2 só dương:
\(\sqrt{x}+\dfrac{1}{\sqrt{x}}\ge2\sqrt{\dfrac{\sqrt{x}.1}{\sqrt{x}}}=2\)
\(\Rightarrow B=1+\sqrt{x}+\dfrac{1}{\sqrt{x}}\ge1+2=3\)
Dấu "=" xảy ra \(\Leftrightarrow x=1\)
